The relative error in area is, to first approximation, twice the relative error in a linear dimension.
That is because
(delta A)/A = (1 + epsilon)^2 - 1
= 1 + 2 epsilon + epsilon^2 - 1
= 2 epsilon
when epsilon << 1
In your case, the answer is
2*0.5/78 = 1/78 = 1.3%
The circumference of a sphere was measured to be 78.000 cm with a possible error of 0.50000 cm. Use linear approximation to estimate the maximum error in the calculated surface area.
Estimate the relative error in the calculated surface area.
1 answer
1 answer